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Let $B=(b_1,b_2,b_3)$ and $C=(c_1,c_2,c_3)$ are two different bases of three dimensional vector space $V$ such that: $$c_1=b_1+3b_2+3b_3$$ $$c_2=4b_2+5b_3$$ $$c_3=2b_3$$

Determine a set of all vectors of $V$ that have the same coordinates in both bases. Check if that set is a vector space and if it is, find its dimension and one basis.

Question: Problem doesn't state what are dimensions of vectors $b_1,b_2,b_3,c_1,c_2,c_3$. How can we find dimension and one basis of a vector space if we don't know the vectors in explicit form?

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Each such vector $x\in V$ corresponds to a number triple $(x_1,x_2,x_3)$ satisfying $$(x=)\quad x_1c_1+x_2c_2+x_3c_3=x_1b_1+x_2b_2+x_3b_3\ .$$ By definition of the $c$-basis $$\eqalign{LHS&=x_1(b_1+3b_2+3b_3)+x_2(4b_2+5b_3)+x_3(2b_3)\cr&=x_1b_1+(3x_1+4x_2)b_2+(3x_1+5x_2+2x_3)b_3\ .\cr}$$ Since $(b_1,b_2,b_3)$ is a basis of $V$ this implies that we have the three equations $$\eqalign{x_1&=x_1\cr3x_1+4x_2&=x_2\cr3x_1+5x_2+2x_3&=x_3\ .\cr}$$ This can be recursively solved as follows: $$x_1={\rm arbitrary},\quad x_2=-x_1,\quad x_3=2x_1\ .$$ The solution set in the triples space ${\mathbb R}^3$ is therefore the set of all triples $\lambda(1,-1,2)$, $\lambda\in{\mathbb R}$, and the set $S$ of all vectors $x\in V$ satisfying the given condition is given by $$S=\{\lambda b_*\,|\,\lambda\in{\mathbb R}\},\qquad b_*=b_1-b_2+2b_3\ ,$$ and is a one-dimensional vector space with basis vector $b_*$.

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  • $\begingroup$ Could you please expand your answer? $\endgroup$ – user300046 Apr 23 '17 at 18:26
  • $\begingroup$ I am getting the following from the LHS and RHS: $x_1(3b_2+3b_3)+x_2(3b_2+5b_3)+x_3b_3=0$. From here, $x_1=x_2=x_3=0$. How to proceed? $\endgroup$ – user300046 Apr 23 '17 at 18:35
  • $\begingroup$ How many element do vectors $b_1,b_2,b_3,c_1,c_2,c_3$ have? Is it relevant here? $\endgroup$ – user300046 Apr 23 '17 at 18:50
  • $\begingroup$ $x_2$ is a free variable, $x_1=-3x_2 ,x_3=4x_2$. This means that the set of all vectors of $V$ that have the same coordinates in both bases is: $S=\{x_2(-3b_1+b_2+4b_3)\}$. Is it correct? $\endgroup$ – user300046 Apr 23 '17 at 19:16
  • $\begingroup$ There is a mistake in my last post. Number triple should be $(-x_2,x_2,-2x_2)$. Set of all vectors of $V$ that have the same coordinates in both bases is $S=\{x_2(-b_1+b_2-2b_3)\}$. How to find its dimension and one basis? $\endgroup$ – user300046 Apr 23 '17 at 20:02
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When you find that $\\(x_1,x_2,x_3)=x_2(-1,1,-2)$ you will know that all the vectors in S can be represented as linear combination of $\{-b_1+b_2-2b_3\} or \{-c_1+c_2-2c_3\} $ and the set of all linear combinations is a span over those vectors. Span is a subspace.

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  • $\begingroup$ Could you show the procedure for finding dimension and one basis of $S$? $\endgroup$ – user300046 Apr 23 '17 at 22:02
  • $\begingroup$ I know the procedure for it when vector coordinates are known, but how to do it here? $\endgroup$ – user300046 Apr 23 '17 at 22:08
  • $\begingroup$ There is a theorem that says If the vectors in span are linear independent they are the basis of that subspace. You have one vector in your span and vector alone is by default independent. Number of vectors in the basis is the dimension. So B={-b1+b2-2b3} is one base of S and dimS=1 $\endgroup$ – CTSnake Apr 23 '17 at 23:11
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What are the "dimensions" of a vector? If you mean the coordinates of the vector, then components are only defined relative to a basis.

You have been given a basis of $V,$ therefore you can express any vector as a list of coordinates over that basis. Actually you have been given two bases, but that just gives you a second way to give coordinates of a vector; it does not prevent you from using the first basis you were given.

So you have $b_1,$ $b_2,$ and $b_3,$ which are a basis. (This is given.) You can express any other basis of the space $V,$ or the basis of any subspace of $V,$ in terms of these three vectors. One such example has already been given in the problem statement: the three vectors $b_1+3b_2+3b_3,$ $4b_2+5b_3,$ and $2b_3$ (also known as $c_1,$ $c_2,$ and $c_3$) are a basis of $V.$ This is a perfectly valid way to write a basis of a vector space, given that we know the vectors $b_1,$ $b_2,$ and $b_3$ are also a basis, and it is the way this question apparently is meant to be answered.

For example, consider the two vectors $b_1$ and $b_1 + b_2 + b_3.$ These vectors are independent, since otherwise you would be able to show that $b_2 + b_3 = 0.$ The span of these two vectors therefore has two dimensions, and because both vectors are vectors in $V,$ their span is a two-dimensional subspace of $V,$ namely, $\{t_1b_1 + t_2(b_1+b_2+b_3)\}.$ The same two vectors ($b_1$ and $b_1 + b_2 + b_3$) are a basis of that subspace.

In case you have not already gotten the hint from this example, when you wrote $S=\{x_2(-b_1+b_2-2b_3)\}$ you had already practically solved the problem. You just need to read off the dimension and the basis from what you wrote.

You do not need to find "coordinates" of $b_1$ or of any of the other given basis vectors of $V$ to solve this problem. The only meaningful coordinates for $b_1$ other than $(1,0,0)$ (defined in terms of the basis $(b_1,b_2,b_3)$) or $\left(1,-\frac34,\frac38\right)$ (defined in terms of the basis $(c_1,c_2,c_3)$) would be defined in terms of some third basis that has not been given--and no such basis has any better claim to be your chosen basis than either $(b_1,b_2,b_3)$ or $(c_1,c_2,c_3).$ So don't look for something you don't have and don't need.

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